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The current president (let's call him Z) has an approval rate of 70% among women and 40% among men. In a dorm with 20 students, 15 of which are women and 5 of which are men, the students are given a form on which they state whether or not they approve of Z. Two of these forms are randomly chosen and both show disapproval for Z. What is the probability that both of these forms were filled in by men?

We name two events:

  • $A: \text{inhabitant is male}$
  • $B: \text{inhabitant disapproves of Z}$

Then we have $$P(A|B) = \dfrac{P(B|A) \cdot P(A)}{P(B)} = \dfrac{0.6 \cdot 0.25}{0.25 \cdot 0.6 + 0.75 \cdot 0.3} = 0.4$$

I believe this is the probability that a randomly drawn form which is negative about Z is from a man. So now we need the probability that the second form which is also negative, is also from a man:

$$P(A|B) = \dfrac{P(B|A) \cdot P(A)}{P(B)} = \dfrac{0.6 \cdot \dfrac{4}{19}}{\dfrac{4}{19} \cdot 0.6 + \dfrac{15}{19} \cdot 0.3} \approx 0.35$$

Then the answer would be $0.4 \cdot 0.35$. I'm feeling somewhat iffy about this answer. Especially the second part, I have a feeling it's incorrect but I don't really know where the mistake lies. Is my approach correct?

Also, I get a different answer if I take the two events and combine them into one event (like the first answer here recommended).

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Do it together. Probability for two dis-approvals, given that both are men = $0.6*0.6=0.36$

Probability that both are men = $\frac{5C_2}{20C_2} = \frac 1{19}$

Probability of two dis-approvals = Probability that both are women $* 0.3*0.3$ + Probability that one is woman and other is man $*0.3*0.6$ + Probability that both are men $*0.6*0.6$

$=>\frac{21}{38}*0.09+\frac{15}{38}*0.18+\frac{2}{19}*0.36$

This will make it quite clear. However, your approach is right as well. You get the same answer.

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  • $\begingroup$ Why $\dfrac{1}{19}$? Isn't it $\dfrac{1}{5}*\dfrac{4}{19}$? $\endgroup$ – GujuratiMC Sep 10 '16 at 15:47
  • $\begingroup$ Also I get different answers for the two methods. My method yields ~14% while combining them yields ~22%. $\endgroup$ – GujuratiMC Sep 10 '16 at 15:50
  • $\begingroup$ It is simply choosing 2 men among 5 divided by total cases. NO It's not $\frac 15 * \frac 4{19}$. It is $\frac 5{20} * \frac 4{19} = \frac 1{19}$ You should get the same answer since you get my equation by multiplying both your initial equations. (Check the denominators for clarity. Multiply your denominators and you get same equation as mine.) $\endgroup$ – Win Vineeth Sep 10 '16 at 16:14
  • $\begingroup$ Oh yeah, $\dfrac{1}{5}$ should be $\dfrac{1}{4}$, brainfart on my part. I still get a different answer though. Hmm.. $\endgroup$ – GujuratiMC Sep 10 '16 at 16:20
  • $\begingroup$ I finally figured it out, thanks. $\endgroup$ – GujuratiMC Sep 10 '16 at 16:33

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